chapter 6: random variables - ms. haralampopoulos · chapter 6 random variables 6.1 discrete and...

+

The Practice of Statistics, 4th edition – For AP*

STARNES, YATES, MOORE

Chapter 6: Random Variables

Section 6.2

Transforming and Combining Random Variables

+ Chapter 6

Random Variables

6.1 Discrete and Continuous Random Variables

6.2 Transforming and Combining Random Variables

6.3 Binomial and Geometric Random Variables

+ Section 6.2


After this section, you should be able to…

DESCRIBE the effect of performing a linear transformation on a

random variable

COMBINE random variables and CALCULATE the resulting mean

and standard deviation

CALCULATE and INTERPRET probabilities involving combinations

of Normal random variables

Learning Objectives

+

Tra

nsfo

rmin

g a

nd

Co

mb

inin

g R

an

do

m V

aria

ble

s

Linear Transformations

In Section 6.1, we learned that the mean and standard deviation give us

important information about a random variable. In this section, we’ll

learn how the mean and standard deviation are affected by

transformations on random variables.

In Chapter 2, we studied the effects of linear transformations on the

shape, center, and spread of a distribution of data. Recall:

1. Adding (or subtracting) a constant, a, to each observation:

• Adds a to measures of center and location.

• Does not change the shape or measures of spread.

2. Multiplying (or dividing) each observation by a constant, b:

• Multiplies (divides) measures of center and location by b.

• Multiplies (divides) measures of spread by |b|.

• Does not change the shape of the distribution.

+ Linear Transformations

Pete’s Jeep Tours offers a popular half-day trip in a tourist area. There

must be at least 2 passengers for the trip to run, and the vehicle will

hold up to 6 passengers. Define X as the number of passengers on a

randomly selected day.

Tra

nsfo

rmin

g a

nd

Co

mb

inin

g R

an

do

m V

aria

ble

s

Passengers xi 2 3 4 5 6

Probability pi 0.15 0.25 0.35 0.20 0.05

The mean of X is 3.75 and the standard

deviation is 1.090.

Pete charges $150 per passenger. The random variable C describes the amount

Pete collects on a randomly selected day.

Collected ci 300 450 600 750 900

Probability pi 0.15 0.25 0.35 0.20 0.05

The mean of C is $562.50 and the standard

deviation is $163.50.

Compare the shape, center, and spread of the two probability distributions.


El Dorado Community College considers a student to be full-time if he

or she is taking between 12 and 18 units. The number of units X that

a randomly selected EDCC full-time student is taking in the fall

semester has the following distribution.

Tra

nsfo

rmin

g a

nd

Co

mb

inin

g R

an

do

m V

aria

ble

s

# of units (X) 12 13 14 15 16 17 18

Probability 0.25 0.10 0.05 0.30 0.10 0.05 .015

The mean of X is 14.65 and the

standard deviation is 2.06.

At EDCC, the tuition for full-time students is $50 per unit. That is, if T=tuition

charge for a randomly selected full-time student, T=50X

Tuition Charge

(T)

$600 $650 $700 $750 $800 $850 $900

Probability 0.25 0.10 0.05 0.30 0.10 0.05 .015


deviation is $103.


Number of Units

Pro

ba

bili

ty

18171615141312

0.30

0.25

0.20

0.15

0.10

0.05

0.00

Tuition Charge

Pro

ba

bili

ty

900850800750700650600

30

25

20

15

10

5

0


How does multiplying or dividing by a constant affect a random

variable?

Tra

nsfo

rmin

g a

nd

Co

mb

inin

g R

an

do

m V

aria

ble

s

Multiplying (or dividing) each value of a random variable by a number b:

• Multiplies (divides) measures of center and location (mean, median,

quartiles, percentiles) by b.

• Multiplies (divides) measures of spread (range, IQR, standard deviation)

by |b|.


Effect on a Random Variable of Multiplying (Dividing) by a Constant

Note: Multiplying a random variable by a constant b multiplies the variance

by b2.


Consider Pete’s Jeep Tours again. We defined C as the amount of

money Pete collects on a randomly selected day.

Tra

nsfo

rmin

g a

nd

Co

mb

inin

g R

an

do

m V

aria

ble

s

It costs Pete $100 per trip to buy permits, gas, and a ferry pass. The random

variable V describes the profit Pete makes on a randomly selected day.

Collected ci 300 450 600 750 900

Probability pi 0.15 0.25 0.35 0.20 0.05




Profit vi 200 350 500 650 800

Probability pi 0.15 0.25 0.35 0.20 0.05

The mean of V is $462.50 and the standard


+ Linear Transformations El Dorado Community College

Tra

nsfo

rmin

g a

nd

Co

mb

inin

g R

an

do

m V

aria

ble

s

Compare the shape, center, and spread

of the two probability distributions.

The mean of V is $832.50 and the standard


Tuition Charge (T) $600 $650 $700 $750 $800 $850 $900

Probability 0.25 0.10 0.05 0.30 0.10 0.05 0.155

Tuition Charge

Pro

ba

bili

ty

900850800750700650600

30

25

20

15

10

5

0


deviation is $103.

In addition to tuition charges, each full-time

student at El Dorado Community College is

assessed student fees of $100 per

semester. If C = overall cost for a randomly

selected full-time student, C = 100 + T.

Tuition Charge (T) $700 $750 $800 $850 $900 $950 $1000

Probability 0.25 0.10 0.05 0.30 0.10 0.05 0.155

Overall Cost

Pro

ba

bili

ty

1000950900850800750700

30

25

20

15

10

5

0


How does adding or subtracting a constant affect a random variable?

Tra

nsfo

rmin

g a

nd

Co

mb

inin

g R

an

do

m V

aria

ble

s

Adding the same number a (which could be negative) to

each value of a random variable:

• Adds a to measures of center and location (mean,

median, quartiles, percentiles).

• Does not change measures of spread (range, IQR,

standard deviation).


Effect on a Random Variable of Adding (or Subtracting) a Constant


Whether we are dealing with data or random variables, the

effects of a linear transformation are the same.

Tra

nsfo

rmin

g a

nd

Co

mb

inin

g R

an

do

m V

aria

ble

s

If Y = a + bX is a linear transformation of the random

variable X, then

• The probability distribution of Y has the same shape

as the probability distribution of X.

• µY = a + bµX.

• σY = |b|σX (since b could be a negative number).

Effect on a Linear Transformation on the Mean and Standard Deviation

+ Alternate Example - Scaling a Test

Problem: In a large introductory statistics class, the distribution of X =

raw scores on a test was approximately normally distributed with a mean

of 17.2 and a standard deviation of 3.8. The professor decides to scale

the scores by multiplying the raw scores by 4 and adding 10.

(a) Define the variable Y to be the scaled score of a randomly selected

student from this class. Find the mean and standard deviation of Y.

(b) What is the probability that a randomly selected student has a scaled

test score of at least 90?

Tra

nsfo

rmin

g a

nd

Co

mb

inin

g R

an

do

m V

aria

ble

s

Solution:

(a) Since Y = 10 + 4X,

and

(b) Since linear transformations do not change

the shape, Y has the N(78.8, 15.2) distribution.

The standardized score for a scaled score of 90

is . According to Table A,

P(z < 0.74) = 0.7704. Thus, P(Y 90) =

1 – 0.7704 = 0.2296.

8.78)2.17(410410 XY

2.15)8.3(44 XY

2.15

8.7890z

+ Combining Random Variables

So far, we have looked at settings that involve a single random variable.

Many interesting statistics problems require us to examine two or

more random variables.

Let’s investigate the result of adding and subtracting random variables.

Let X = the number of passengers on a randomly selected trip with

Pete’s Jeep Tours. Y = the number of passengers on a randomly

selected trip with Erin’s Adventures. Define T = X + Y. What are the

mean and variance of T?

Tra

nsfo

rmin

g a

nd

Co

mb

inin

g R

an

do

m V

aria

ble

s

Passengers xi 2 3 4 5 6

Probability pi 0.15 0.25 0.35 0.20 0.05

Passengers yi 2 3 4 5

Probability pi 0.3 0.4 0.2 0.1

Mean µX = 3.75 Standard Deviation σX = 1.090

Mean µY = 3.10 Standard Deviation σY = 0.943


How many total passengers can Pete and Erin expect on a

randomly selected day?

Since Pete expects µX = 3.75 and Erin expects µY = 3.10 , they

will average a total of 3.75 + 3.10 = 6.85 passengers per trip.

We can generalize this result as follows:

Tra

nsfo

rmin

g a

nd

Co

mb

inin

g R

an

do

m V

aria

ble

s

For any two random variables X and Y, if T = X + Y, then the

expected value of T is

E(T) = µT = µX + µY

In general, the mean of the sum of several random variables is the

sum of their means.

Mean of the Sum of Random Variables

How much variability is there in the total number of passengers who

go on Pete’s and Erin’s tours on a randomly selected day? To

determine this, we need to find the probability distribution of T.

+ Alternate Example – El Dorado Community College

El Dorado Community College also has a campus downtown,

specializing in just a few fields of study. Full-time students at

the downtown campus only take 3-unit classes. Let Y =

number of units taken in the fall semester by a randomly

selected full-time student at the downtown campus. Here is

the probability distribution of Y:

Tra

nsfo

rmin

g a

nd

Co

mb

inin

g R

an

do

m V

aria

ble

s

The mean of this distribution is = 15 units, the variance is

= 5.40 units2 and the standard deviation is = 2.3 units.

If you were to randomly select 1 full-time student from the main

campus and 1 full-time student from the downtown campus and add

their number of units, the expected value of the sum (S = X + Y)

would be:

.

Number of Units (Y) 12 15 18

Probability 0.3 0.4 0.3

65.291565.14 yxs

Y2

Y Y


The only way to determine the probability for any value of T is if X and Y

are independent random variables.

Tra

nsfo

rmin

g a

nd

Co

mb

inin

g R

an

do

m V

aria

ble

s

Definition:

If knowing whether any event involving X alone has occurred tells us

nothing about the occurrence of any event involving Y alone, and vice

versa, then X and Y are independent random variables.

Probability models often assume independence when the random variables

describe outcomes that appear unrelated to each other.

You should always ask whether the assumption of independence seems

reasonable.

In our investigation, it is reasonable to assume X and Y are independent

since the siblings operate their tours in different parts of the country.


Let T = X + Y. Consider all possible combinations of the values of X and Y.

Recall: µT = µX + µY = 6.85

T2 (ti T )2 pi

= (4 – 6.85)2(0.045) + … +

(11 – 6.85)2(0.005) = 2.0775

Note:

X2 1.1875 and Y

2 0.89

What do you notice about the

variance of T?

+ Combining Random Variables – El Dorado Community College

Let S = X + Y as before. Assume that X and Y are independent, which is

reasonable since each student was selected at random. Here are all

possible combinations of X and Y.

µS = 24(0.075)+25(0.03)+…

+36(0.045)=29.65

63.9)045.0()65.2936(

...)03.0()65.2925(

)075.0()65.2924(

2

2

22

S

Notice that

)1565.1465.29( YXs

and that

X P(X) Y P(Y) S=X+Y P(S)=P(X

)P(Y)

12 0.25 12 0.3 24 0.075

12 0.25 15 0.4 27 0.10

12 0.25 18 0.3 30 0.075

13 0.10 12 0.3 25 0.03

13 0.10 15 0.4 28 0.04

13 0.10 18 0.3 31 0.03

14 0.05 12 0.3 26 0.015

14 0.05 15 0.4 29 0.02

14 0.05 18 0.3 32 0.015

15 0.30 12 0.3 27 0.09

15 0.30 15 0.4 30 0.12

15 0.30 18 0.3 33 0.09

16 0.10 12 0.3 28 0.03

16 0.10 15 0.4 31 0.04

16 0.10 18 0.3 34 0.03

17 0.05 12 0.3 29 0.015

17 0.05 15 0.4 32 0.02

17 0.05 18 0.3 35 0.015

18 0.15 12 0.3 30 0.045

18 0.15 15 0.4 33 0.06

18 0.15 18 0.3 36 0.045

S 24 25 26 27 28 29 30 31 32 33 34 35 36

P(S) 0.075 0.03 0.015 0.19 0.07 0.035 0.24 0.07 0.035 0.15 0.03 0.015 0.045

Here is the probability distribution of S:

)40.523.463.9(222 YXS


As the preceding example illustrates, when we add two

independent random variables, their variances add. Standard

deviations do not add.

Tra

nsfo

rmin

g a

nd

Co

mb

inin

g R

an

do

m V

aria

ble

s

Variance of the Sum of Random Variables

Remember that you can add variances only if the two random variables are

independent, and that you can NEVER add standard deviations!

For any two independent random variables X and Y, if T = X + Y, then the

variance of T is

In general, the variance of the sum of several independent random variables

is the sum of their variances.

T2 X

2 Y2

+ Alternate Example – Tuition, Fees, and Books

Let B = the amount spent on books in the fall semester for a

randomly selected full-time student at El Dorado Community

College. Suppose that and . Recall from

earlier that C = overall cost for tuition and fees for a randomly

selected full-time student at El Dorado Community College

and = $832.50 and = $103.

Problem: Find the mean and standard deviation of the cost of

tuition, fees, and books (C + B) for a randomly selected full-

time student at El Dorado Community College.

Tra

nsfo

rmin

g a

nd

Co

mb

inin

g R

an

do

m V

aria

ble

s

153B 32B

C C

Solution: The mean is = 832.50 + 153 = $985.50.

The standard deviation cannot be calculated since the cost for

tuition and fees and the cost for books are not independent.

Students who take more units will typically have to buy more

books.

BCBC


Problem:

(a) At the downtown campus, full-time students pay $55 per unit.

Let U = cost of tuition for a randomly selected full-time student at

the downtown campus. Find the mean and standard deviation of U.

(b) Calculate the mean and standard deviation of the total amount

of tuition for a randomly selected full-time student at the main

campus and for a randomly selected full-time student at the

downtown campus.

Tra

nsfo

rmin

g a

nd

Co

mb

inin

g R

an

do

m V

aria

ble

s

Solution:

(a)

(b)

thus

50.126$)3.2(5555

,825$)15(55

YU

U

570,26002,16568,10

.50.155782550.732

222

UTUT

UTUT

163$570,26 UT


We can perform a similar investigation to determine what happens

when we define a random variable as the difference of two random

variables. In summary, we find the following:

Tra

nsfo

rmin

g a

nd

Co

mb

inin

g R

an

do

m V

aria

ble

s

Variance of the Difference of Random Variables

For any two independent random variables X and Y, if D = X - Y, then the

variance of D is

In general, the variance of the difference of two independent random

variables is the sum of their variances.

D2 X

2 Y2

For any two random variables X and Y, if D = X - Y, then the expected value

of D is

E(D) = µD = µX - µY

In general, the mean of the difference of several random variables is the

difference of their means. The order of subtraction is important!

Mean of the Difference of Random Variables


Problem: Suppose we randomly select one full-time student from

each of the two campuses. What are the mean and standard

deviation of the difference in tuition charges, D = T – U? Interpret

each of these values.

Tra

nsfo

rmin

g a

nd

Co

mb

inin

g R

an

do

m V

aria

ble

s

Solution:

;570,26002,16568,10222 UTUT

.50.9282550.732 UTUT

163$570,26 UT

This means that, on average, full-time students at the main campus

pay $92.50 less in tuition than full-time students at the downtown

campus.

thus

Although the average difference in tuition for the two campuses is –

$92.50, the difference in tuition for a randomly selected full-time

student from each college will vary from the average difference by

about $163, on average. Notice that the standard deviation is the

same for the sum of tuition costs and the difference of tuition costs.

+ Combining Normal Random Variables

So far, we have concentrated on finding rules for means and variances

of random variables. If a random variable is Normally distributed, we

can use its mean and standard deviation to compute probabilities.

An important fact about Normal random variables is that any sum or

difference of independent Normal random variables is also Normally

distributed.

Tra

nsfo

rmin

g a

nd

Co

mb

inin

g R

an

do

m V

aria

ble

s

Mr. Starnes likes between 8.5 and 9 grams of sugar in his hot tea. Suppose

the amount of sugar in a randomly selected packet follows a Normal distribution

with mean 2.17 g and standard deviation 0.08 g. If Mr. Starnes selects 4 packets

at random, what is the probability his tea will taste right?

Let X = the amount of sugar in a randomly selected packet.

Then, T = X1 + X2 + X3 + X4. We want to find P(8.5 ≤ T ≤ 9).

µT = µX1 + µX2 + µX3 + µX4 = 2.17 + 2.17 + 2.17 +2.17 = 8.68

T2 X1

2 X2

2 X3

2 X4

2 (0.08)2 (0.08)2 (0.08)2 (0.08)2 0.0256

T 0.0256 0.16

P(-1.13 ≤ Z ≤ 2.00) = 0.9772 – 0.1292 = 0.8480

There is about an 85% chance Mr. Starnes’s

tea will taste right.

z 8.5 8.68

0.16 1.13 and z

9 8.68

0.16 2.00

Example

+ Alternate Example - Apples

Suppose that the weights of a certain variety of apples have weights that are

N(9,1.5). If bags of apples are filled by randomly selecting 12 apples, what is the

probability that the sum of the weights of the 12 apples is less than 100 ounces?

State: What is the probability that a random sample of 12 apples has a total

weight less than 100 ounces?

Plan: Let X = weight of a randomly selected apple. Then X1 = weight of first

randomly selected apple, etc. We are interested in the total weight

T = X1 + X2 + + X12. Our goal is to find P(T < 100).

Do: Since T is a sum of 12 independent Normal random variables, T follows a

Normal distribution with mean µT = µX1 + µX2 +… + µX12 = 9 + 9 + … + 9 = 108

ounces and variance

The standard deviation is = 5.2 ounces.

P(T < 100) = normalcdf(–99999, 100, 108, 5.2) = 0.0620. Note: to get full credit

on the AP exam when using the calculator command normalcdf, students must

clearly identify the shape (Normal), center (mean = 108) and spread (standard

deviation = 5.2) somewhere in their work.

Conclude: There is about a 6.2% chance that the 12 randomly selected apples

will have a total weight of less than 100 ounces.

Tra

nsfo

rmin

g a

nd

Co

mb

inin

g R

an

do

m V

aria

ble

s

275.1...5.15.1... 2222

12

222

21 XXXT

27T

+

Alternate Example – Speed Dating

Suppose that the height M of male speed daters follows is N(70, 3.5) and the

height F of female speed daters follows a Normal distribution with a mean of 65

inches and a standard deviation of 3 inches. What is the probability that a

randomly selected male speed dater is taller than the randomly selected female

speed dater with whom he is paired?

State: What is the probability that a randomly selected male speed dater is taller

than the randomly selected female speed dater with whom he is paired?

Plan: We’ll define the random variable D = M – F to represent the difference

between the male’s height and the female’s height. Our goal is to find P(M > F)

or P(D > 0).

Do: Since D is the difference of two independent Normal random variables, D

follows a Normal distribution with mean µD = µM - µF = 70 – 65 = 5 inches and

variance . Thus, the standard deviation is

inches.

Thus, P(D > 0) = normalcdf(0, 99999, 5, 4.61) = 0.8610. Note: to get full credit

on the AP exam when using the calculator command normalcdf, students must

clearly identify the shape (Normal), center (mean = 5) and spread (standard

deviation = 4.61) somewhere in their work.

Conclude: There is about an 86% chance that a randomly selected male speed

dater will be taller than the female he is randomly paired with. Or, in about 86%

of speed dating couples, the male will be taller than the female.

Tra

nsfo

rmin

g a

nd

Co

mb

inin

g R

an

do

m V

aria

ble

s

25.2135.3 2222 FXD M

61.425.21 D

+ Section 6.2


In this section, we learned that…

Adding a constant a (which could be negative) to a random variable

increases (or decreases) the mean of the random variable by a but does not

affect its standard deviation or the shape of its probability distribution.

Multiplying a random variable by a constant b (which could be negative)

multiplies the mean of the random variable by b and the standard deviation

by |b| but does not change the shape of its probability distribution.

A linear transformation of a random variable involves adding a constant a,

multiplying by a constant b, or both. If we write the linear transformation of X

in the form Y = a + bX, the following about are true about Y:

Shape: same as the probability distribution of X.

Center: µY = a + bµX

Spread: σY = |b|σX

Summary

+ Section 6.2


In this section, we learned that…

If X and Y are any two random variables,

If X and Y are independent random variables

The sum or difference of independent Normal random variables follows a

Normal distribution.

Summary

X Y2 X

2 Y2

X Y X Y

+ Looking Ahead…

We’ll learn about two commonly occurring discrete random

variables: binomial random variables and geometric

random variables.

We’ll learn about

Binomial Settings and Binomial Random Variables

Binomial Probabilities

Mean and Standard Deviation of a Binomial

Distribution

Binomial Distributions in Statistical Sampling

Geometric Random Variables

In the next Section…

chapter 6: random variables - ms. haralampopoulos · chapter 6 random variables 6.1 discrete and...

Documents